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The problem consists of using the epsilon neighborhood definition to prove that A\B is open and B\A is closed.

I'm thinking of using a proof by contradiction and assuming that A\B is not open and vice versa for B\A. Any help would be greatly appreciated.

Sorry, forgot to mention that we can't use the fact that a set is open iff its compliment is closed and vice versa.

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  • $\begingroup$ Do you mean $A \setminus B$ and $B \setminus A$? Otherwise, I'm not sure what you mean by A/B or B/A. $\endgroup$ – Daniel Hast Oct 22 '19 at 21:27
  • $\begingroup$ Yes, sorry about that; meant the difference of sets. $\endgroup$ – carbonv2 Oct 22 '19 at 21:28
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    $\begingroup$ How do you define "closed" then? $\endgroup$ – amsmath Oct 22 '19 at 21:39
  • $\begingroup$ If the limit points of the set are contained in the set. $\endgroup$ – carbonv2 Oct 22 '19 at 21:41
  • $\begingroup$ "Sorry, forgot to mention that we can't use the fact that a set is open iff its compliment is closed and vice versa." ... well, just echo that proof of that.... $\endgroup$ – fleablood Oct 22 '19 at 22:29
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$A$ is open and $B$ is closed. Let $x \in A \setminus B$. Then there is some $r>0$ such that $B(x,r) \subseteq A$ as $A$ is open and $x \in A$.

$x \notin B$ so $x$ is not a limit point of $B$ (as $B$ is closed it would contain all its limit points), so there is soem $s>0$ such that $B(x,s) \cap B =\emptyset$.

But then $B(x, \min(r,s)) \subseteq A\setminus B$ showing $x$ to be an interior point of $A \setminus B$, so $A \setminus B$ is open.

Now suppose $x$ is a limit point of $B\setminus A$. Then $x$ is also a limit point of the larger set $B$ and as $B$ is closed, $x \in B$. Suppose that $x \in A$, then we'd have $r>0$ such that $B(x,r) \subseteq A$ and then $B(x,r) \cap (B \setminus A)=\emptyset$, and $x$ wouldn't be a limit point of $B \setminus A$. So, in fact, we must have that $x \notin A$, which together with $x \in B$ implies $x \in B\setminus A$, which thus contains all its limit points and is thus closed.

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A faster solution If $B$ is closed, its complementary $C$ is open, $A-B=A\cap C$ is open as the intersection of two open sets.

The complementary $D$ of $A$ is closed and $B-A=B\cap D$.

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$$A\setminus B=A\cap B^c$$ Note that $B^c$ is open and $A^c$ is closed. The result naturally follows.

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  • $\begingroup$ Why the downvote? $\endgroup$ – Don Thousand Oct 22 '19 at 21:25
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    $\begingroup$ The original poster is explicitly referring to the epsilon-neighbourhood definition and implicitly to real analysis (one of the tags). Your method of proof may constitute a valid answer to his question - or it may not. $\endgroup$ – Alexander Geldhof Oct 22 '19 at 21:26
  • $\begingroup$ @AlexanderGeldhof It also has the tag general-topology. $\endgroup$ – Don Thousand Oct 22 '19 at 21:27
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    $\begingroup$ Sure, I'll let OP decide. Your proof is obviously mathematically valid, but I believe they may be looking for a different way of proof. EDIT: the edit to their post seems to imply I'm right. $\endgroup$ – Alexander Geldhof Oct 22 '19 at 21:28
  • $\begingroup$ Alexander is correct; I forgot to mention that I cannot use the compliments theorem and can only use the epsilon neighborhood definition. Thanks for helping me clarify. $\endgroup$ – carbonv2 Oct 22 '19 at 21:30
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Let $x\in A\setminus B$. Denote by $U_\epsilon$ the $\epsilon$-neighborhood of $x$. If for any $n$ there was a point $x_n\in B\cap U_{1/n}$, then $x_n\to x$. But $B$ is closed and so $x\in B$. Contradiction! Hence, there exists $n$ such that $U_{1/n}$ is completely contained in $A\setminus B$.

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Just use definitions.

Let $k$ be a limit point of $B\setminus A$. Then $k$ is a limit point of $B$ as $B\setminus A \subset B$. (Obviously if every neighborhood of $k$ has a point $x$not equal to $k$ so that $x\in B\setminus A$ then $x \in B$.) So $k\in B$.

And if $k \in A$ then as $A$ is open there is a radius $r$ so that $B_r(k) \subset A$. But then $B_r(k)$ has no points of $B\setminus A$. But this contradicts $k$ is a limit point of $B\setminus A$.

So $k \in B\setminus A$. So all limit points of $B\setminus A$ are in $B\setminus A$ and so $B\setminus A$ is closed.

....

And let $a \in A\setminus B$ then $a \in A$ and as $A$ is open then there is a $r_1$ so that $B_{r_1}(a)\subset A$. And $a \not \in B$ and $B$ is closed so $a$ is not a limit point of $B$. So there is an $r_2$ so that $B_{r_2}(a) \cap B = \emptyset$.

So $B_{\min(r_2,r_1)}\subset B_{r_1}(a) \subset A$ and $B_{\min(r_2,r_1)}\subset B_{r_2}(a)\subset B^c$. So $B_{\min(r_2,r_1)} \subset A\cap B^c = A\setminus B$.

And so $A\setminus B$ is open.

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If we assume A/B is not open, then there must be a point $x$ in $A \setminus B$ for which all neighbourhoods intersect $(A \setminus B)^c$. $(A \setminus B)^c$ is the set of all points which either lie in $B$, or lie not in $A$ (or both).

Pick a point in $A \setminus B$. It lies in $A$, so at least one of its neighbourhoods lies completely in $A$. It lies in $B^c$. There has to be at least one of its neighbourhoods which lies completely in $B^c$ (i.e. does not intersect $B$), as $B$ is its own closure.

Take the intersection of it with the first neighbourhood to provide a counterexample for our assumption that $A \setminus B$ is not open.

A similar idea could be used for the $B \setminus A$ case.

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